Twitter-Bootstrap modal 中的简单 Django 表单

Question

我正在尝试使用Twitter-Bootstrap模式运行django表单。我想知道提交表单后我应该怎么做才能返回/。我的views和templates如下。

url.py

urlpatterns = patterns('myapp.views',
    url(r'^$', 'main'),
    url(r'^add/', 'form_add'),
)

views.py

def main(request):
    if request.method == 'POST':
        form = MyModelForm(request.POST)
        if form.is_valid():
            name = form.cleaned_data['name']
            request.session['name'] = name
            mm = MyModel.objects.create(name=name)
            mm.save()
            return HttpResponseRedirect('/add') # Redirect after POST
    else:
        form = MyModelForm()
    args = {}
    args['last_item'] = MyModel.objects.all().order_by('pk').reverse()[0]
    args['form'] = form
    return render(request, 'form.html', args)
def form_add(request):
    args = {}
    name = request.session['name']
    return render(request, 'add.html', args)

形式.html

{% extends "base.html" %}
{% block content %}
<button type="button" data-toggle="modal"
        data-target="#myModal1">Launch modal</button>
<div class="modal" id="myModal1" tabindex="-1" role="dialog"
     aria-labelledby="myModal1Label" aria-hidden="true" style="display: none">
  <div class="modal-header">
    <button type="button" class="close" data-dismiss="modal" aria-hidden="true">×</button>
    <h3 id="myModal1Label">Modal header</h3>
  </div>
  <div class="modal-body">
    <form method="POST" id="" action="">
      {% csrf_token %}
      {{ form.as_p }}
      <button>Submit</button>
    </form>
  </div>
</div>
<p>Last item: {{ last_item }}</p>
{% endblock %}
{% block scripts %}
{% endblock %}

Answer 1

注释转换为答案。

HttpResponseRedirect('/')而不是'/add'？